Consider the indefinite integral
∫
4
x
3
+
5
x
2
−
96
x
−
125
x
2
−
25
,
d
x
\displaystyle \int{\frac{{{4}{x}^{{3}}+{5}{x}^{{2}}-{96}{x}-{125}}}{{{x}^{{2}}-{25}}}},{\left.{d}{x}\right.}
∫
x
2
−
25
4
x
3
+
5
x
2
−
96
x
−
125
,
d
x
Then the integrand decomposes into the form
a
x
+
b
+
c
x
−
5
+
d
x
+
5
\displaystyle {a}{x}+{b}+{\frac{{{c}}}{{{x}-{5}}}}+{\frac{{{d}}}{{{x}+{5}}}}
a
x
+
b
+
x
−
5
c
+
x
+
5
d
where
a
\displaystyle {a}
a
=
b
\displaystyle {b}
b
=
c
\displaystyle {c}
c
=
d
\displaystyle {d}
d
=
Integrating term by term, we obtain that
∫
4
x
3
+
5
x
2
−
96
x
−
125
x
2
−
25
,
d
x
=
\displaystyle \int{\frac{{{4}{x}^{{3}}+{5}{x}^{{2}}-{96}{x}-{125}}}{{{x}^{{2}}-{25}}}},{\left.{d}{x}\right.}=
∫
x
2
−
25
4
x
3
+
5
x
2
−
96
x
−
125
,
d
x
=
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Question 6 Part 5 of 5
+
C
\displaystyle +{C}
+
C
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